June 2006 6679 Mechanics M3 Mark Scheme Question Number
Scheme
1.
Use of y 2 dx x xy 2 dx
Marks
x dx x x
2
M1
dx
1 2 1 3 2 x x 3 x 16 64 x 2 3 8 x 3
Using limits 0 and 4
A1 = A1 M1 A1
(5) [5]
2.
(a)
Small Hemisphere Mass ratios
2 a 3 2
x
3 a 16
3
Bowl
2 7a 3 8
Large Hemisphere 3
2 3 a 3 Anything in the ratio 1 : 7 : 8 3 a 8
x
3 3 a 7 x 8 a 16 8 45 Leading to x a 112 1
(b)
Bowl M
Mass Ratios
45 a 112
x
M
Liquid kM
Bowl and Liquid k 1 M
3 a 16
17 a 48
45 3 17 a kM a k 1 M a 112 16 48 2 Leading to k 7
B1
B1
M1 A1 cso
A1
(5)
B1 B1
M1 A1 A1
(5) [10]
1
Question Number
3.
Scheme
Marks
a 0.1 2 1 10 5 Fmax ma 2
(a)
B1 M1 A1 M1
0.2 0.1 10
2
N
19.7
A1 (6)
cao a 0.2 ,
(b)
10
v2 2 a 2 x2 100 2 0.22 0.12
v 5.44
3 ms
2
B1ft, B1ft
29.6 ...
1
cao If answers are given to more than 3 significant figures a maximum of one A mark is lost in the question.
4.
M1
tan
3 4
tan
r h
M1 A1 A1 (5) [11]
B1
or equivalent R r 3a or h 4a
R
r 5 R mg 3 h
mr
B1
R sin mg
M1 A1
R
mg
R cos mr 2
8g 10mrg R 9a 9a
tan
M1 A1 A1
9a 5 10mrg mg 8r 3 9a
M1 A1
Eliminating R 3 3 9a r a 2 4 8r r 3a 4 h 2a tan 2 3
2
M1 A1 (11)
[11]
3
Question Number
5.
Scheme
(a)
Marks
A 0.75 m
B 1m
P AP 0.752 12 = 1.25 Conservation of energy 1 49 0.52 2 v2 2 2 g 1 1 2 2 0.75 for each incorrect term Leading to v 1.8 ms 1
M1 A1
M1 A2 (1, 0)
A1
(6)
accept 1.81 (b) A 0.75 m
y T
B
T
P 2g R
2T cos 2 g
0.75 sin 49 0.75 T 0.75 0.75 sin 1 49 1 sin 9.8 1 49 1 cos sin
M1 A1
y
Hooke’s Law
Eliminating T
tan 5 1 sin 5 tan 5sin
cso
M1 A1
M1
A1
(6) [12]
4
5
Question Number
6.
Scheme
Marks
(a) v B1
Parabola 15
B1
Hyperbola
B1 (3)
Points 7.5
O
4
5
10 t
(b) Identifying the minimum point of the parabola and 5 as the end points.
A1 (2)
2t 5
(c) Splitting the integral into two part, with limits 0 and 4, and 4 and 5, and evaluating both integrals.
4 0
3t t 4 dt t 3 6t 2 32 and 4
0
5 4
3t t 4 dt t 3 6t 2 7
Both Total distance 39
5
4
m
cso
(d)
t1 5
75 dt 32 7 t
75 ln t 51 25 t
ln
1 t1 1 t1 5e 3 5 3 6.98
cao
M1
M1 A1 A1 (3)
M1 A1 A1 M1 A1 (5) [13]
6
7
Question Number
7.
Scheme
Marks
(a) A
u
P
5 gl 2
Conservation of Energy 1 5 gl m u 2 mgl 2 2 gl Leading to u 2
M1 A1= A1
(b)
A1
(4)
v
T
mg
u A
B
Conservation of Energy 1 m u 2 v 2 mgr 2 v 2 u 2 2 gr R T
T mg
mv 2 r
m 2 u 2 gr mg r mu 2 3mg r mgl 3mg 2r mgl 3mg 2r 1 r 6 5l ABMIN 6
T 0
r
M1 A1
M1 A1 M1 A1 M1
M1
A1
(9) [13]
8
9
10
11